# How do you factor y= 2x^3-2x^2-4x ?

Dec 26, 2015

$y = 2 {x}^{3} - 2 {x}^{2} - 4 x = 2 x \left(x + 1\right) \left(x - 2\right)$

#### Explanation:

First note that all of the terms are divisible by $2 x$, so separate that out as a factor:

$y = 2 {x}^{3} - 2 {x}^{2} - 4 x = 2 x \left({x}^{2} - x - 2\right)$

Then notice that if you substitute $x = - 1$ into ${x}^{2} - x - 2$, the result is $0$, so $\left(x + 1\right)$ is a factor and the other factor must be $\left(x - 2\right)$ to get the $- 2$ constant term when multiplied:

$\left({x}^{2} - x - 2\right) = \left(x + 1\right) \left(x - 2\right)$

So we have:

$y = 2 {x}^{3} - 2 {x}^{2} - 4 x = 2 x \left(x + 1\right) \left(x - 2\right)$